Chapter 17: Additional Aspects of Acid-Base Equilibria Chemistry 140 Fall 2002 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci • Harwood • Herring • Madura Chapter 17: Additional Aspects of Acid-Base Equilibria Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 General Chemistry: Chapter 17 Prentice-Hall © 2007

Contents 17-1 The Common-Ion Effect in Acid-Base Equilibria Chemistry 140 Fall 2002 Contents 17-1 The Common-Ion Effect in Acid-Base Equilibria 17-2 Buffer Solutions 17-3 Acid-Base Indicators 17-4 Neutralization Reactions and Titration Curves 17-5 Solutions of Salts of Polyprotic Acids 17-6 Acid-Base Equilibrium Calculations: A Summary Focus On Buffers in Blood General Chemistry: Chapter 17 Prentice-Hall © 2007

17-1 The Common-Ion Effect in Acid-Base Equilibria The Common-Ion Effect describes the effect on an equilibrium by a second substance that furnishes ions that can participate in that equilibrium. The added ions are said to be common to the equilibrium. General Chemistry: Chapter 17 Prentice-Hall © 2007

Solutions of Weak Acids and Strong Acids Consider a solution that contains both 0.100 M CH3CO2H and 0.100 M HCl. CH3CO2H + H2O CH3CO2- + H3O+ (0.100-x) M x M x M HCl + H2O Cl- + H3O+ 0.100 M 0.100 M [H3O+] = (0.100 + x) M essentially all due to HCl General Chemistry: Chapter 17 Prentice-Hall © 2007

Acetic Acid and Hydrochloric Acid 0.1 M HCl + 0.1 M CH3CO2H 0.1 M CH3CO2H 0.1 M CH3CO2H + 0.1 M CH3CO2Na General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-1 Demonstrating the Common-Ion Effect: Solution of a weak Acid and a Strong Acid. (a) Determine [H3O+] and [CH3CO2-] in 0.100 M CH3CO2H. (b) Then determine these same quantities in a solution that is 0.100 M in both CH3CO2H and HCl. Recall Example 17-6 (p 680): CH3CO2H + H2O → H3O+ + CH3CO2- [H3O+] = [CH3CO2-] = 1.310-3 M General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-1 CH3CO2H + H2O → H3O+ + CH3CO2- Initial concs. weak acid 0.100 M 0 M 0 M strong acid 0 M 0.100 M 0 M Changes -x M +x M +x M Equilibrium (0.100 - x) M (0.100 + x) M x M Concentration Assume x << 0.100 M, 0.100 – x  0.100 + x  0.100 M General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-1 CH3CO2H + H2O → H3O+ + CH3CO2- Eqlbrm conc. (0.100 - x) M (0.100 + x) M x M Assume x << 0.100 M, 0.100 – x  0.100 + x  0.100 M [H3O+] [CH3CO2-] [C3CO2H] Ka= x · (0.100 + x) (0.100 - x) = x · (0.100) (0.100) = 1.810-5 [CH3CO2-] = 1.810-5 M compared to 1.310-3 M. Le Châtelier’s Principle General Chemistry: Chapter 17 Prentice-Hall © 2007

Suppression of Ionization of a Weak Acid General Chemistry: Chapter 17 Prentice-Hall © 2007

Suppression of Ionization of a Weak Base General Chemistry: Chapter 17 Prentice-Hall © 2007

Solutions of Weak Acids and Their Salts General Chemistry: Chapter 17 Prentice-Hall © 2007

Solutions of Weak Bases and Their Salts General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 17-2 Buffer Solutions Two component systems that change pH only slightly on addition of acid or base. The two components must not neutralize each other but must neutralize strong acids and bases. A weak acid and it’s conjugate base. A weak base and it’s conjugate acid General Chemistry: Chapter 17 Prentice-Hall © 2007

Pure Water Has No Buffering Ability General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 Buffer Solutions Consider [CH3CO2H] = [CH3CO2-] in a solution. [H3O+] [CH3CO2-] Ka= = 1.810-5 [C3CO2H] [CH3CO2-] [C3CO2H] Ka [H3O+] = = 1.810-5 pH = -log[H3O+] = -logKa = -log(1.810-5) = 4.74 General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 How A Buffer Works General Chemistry: Chapter 17 Prentice-Hall © 2007

Preparing a Buffer Solution General Chemistry: Chapter 17 Prentice-Hall © 2007

The Henderson-Hasselbalch Equation A variation of the ionization constant expression. Consider a hypothetical weak acid, HA, and its salt NaA: HA + H2O A- + H3O+ [H3O+] [A-] [HA] Ka= [H3O+] [HA] Ka= [A-] -log[H3O+]-log [HA] -logKa= [A-] General Chemistry: Chapter 17 Prentice-Hall © 2007

Henderson-Hasselbalch Equation -log[H3O+] - log [HA] -logKa= [A-] pH - log [HA] pKa = [A-] pKa + log [HA] pH = [A-] pKa + log [acid] pH = [conjugate base] General Chemistry: Chapter 17 Prentice-Hall © 2007

Henderson-Hasselbalch Equation Chemistry 140 Fall 2002 Henderson-Hasselbalch Equation pKa + log [acid] pH= [conjugate base] Only useful when you can use initial concentrations of acid and salt. This limits the validity of the equation. Limits can be met by: [A-] 0.1 < < 10 [HA] [A-] > 10Ka and [HA] > 10Ka General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-5 Preparing a Buffer Solution of a Desired pH. What mass of NaC2H3O2 must be dissolved in 0.300 L of 0.25 M HC2H3O2 to produce a solution with pH = 5.09? (Assume that the solution volume is constant at 0.300 L) Equilibrium expression: HC2H3O2 + H2O C2H3O2- + H3O+ [H3O+] [HC2H3O2] Ka= [C2H3O2-] = 1.810-5 General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-5 [C2H3O2-] Ka= [H3O+] = 1.810-5 [HC2H3O2] [H3O+] = 10-5.09 = 8.110-6 [HC2H3O2] = 0.25 M Solve for [C2H3O2-] [H3O+] [HC2H3O2] = Ka [C2H3O2-] = 0.56 M 8.110-6 0.25 = 1.810-5 General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-5 [C2H3O2-] = 0.56 M 0.56 mol 1 mol NaC2H3O2 mass C2H3O2- = 0.300 L   1 L 1 mol C2H3O2- 82.0 g NaC2H3O2  = 14 g NaC2H3O2 1 mol NaC2H3O2 General Chemistry: Chapter 17 Prentice-Hall © 2007

Six Methods of Preparing Buffer Solutions General Chemistry: Chapter 17 Prentice-Hall © 2007

Calculating Changes in Buffer Solutions General Chemistry: Chapter 17 Prentice-Hall © 2007

Buffer Capacity and Range Buffer capacity is the amount of acid or base that a buffer can neutralize before its pH changes appreciably. Maximum buffer capacity exists when [HA] and [A-] are large and approximately equal to each other. Buffer range is the pH range over which a buffer effectively neutralizes added acids and bases. Practically, range is 2 pH units around pKa. General Chemistry: Chapter 17 Prentice-Hall © 2007

17-3 Acid-Base Indicators Color of some substances depends on the pH. HIn + H2O In- + H3O+ In the acid form the color appears to be the acid color. In the base form the color appears to be the base color. Intermediate color is seen in between these two states. The complete color change occurs over about 2 pH units. General Chemistry: Chapter 17 Prentice-Hall © 2007

Indicator Colors and Ranges Slide 27 of 45 General Chemistry: Chapter 17 Prentice-Hall © 2007 General Chemistry: Chapter 17 Prentice-Hall © 2007

Testing the pH of a Swimming Pool General Chemistry: Chapter 17 Prentice-Hall © 2007

17-4 Neutralization Reactions and Titration Curves Equivalence point: The point in the reaction at which both acid and base have been consumed. Neither acid nor base is present in excess. End point: The point at which the indicator changes color. Titrant: The known solution added to the solution of unknown concentration. Titration Curve: The plot of pH vs. volume. General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 The millimole Typically: Volume of titrant added is less than 50 mL. Concentration of titrant is less than 1 mol/L. Titration uses less than 1/1000 mole of acid and base. L/1000 mol/1000 = M = L mol mL mmol General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Strong Acid with a Strong Base General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Strong Acid with a Strong Base The pH has a low value at the beginning. The pH changes slowly: until just before the equivalence point. The pH rises sharply: perhaps 6 units per 0.1 mL addition of titrant. The pH rises slowly again. Any Acid-Base Indicator will do. As long as color change occurs between pH 4 and 10. General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Strong Base with a Strong Acid General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Weak Acid with a Strong Base General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Weak Acid with a Strong Base General Chemistry: Chapter 17 Prentice-Hall © 2007

Titration of a Weak Polyprotic Acid NaOH NaOH H3PO4 H2PO4- HPO42-  PO43- General Chemistry: Chapter 17 Prentice-Hall © 2007

17-5 Solutions of Salts of Polyprotic Acids The third equivalence point of phosphoric acid can only be reached in a strongly basic solution. The pH of this third equivalence point is not difficult to calculate. It corresponds to that of Na3PO4 (aq) and PO43- can ionize only as a base. PO43- + H2O → OH- + HPO42- Kb = Kw/Ka = 2.410-2 General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-9 Determining the pH of a Solution Containing the Anion (An-) of a Polyprotic Acid. Sodium phosphate, Na3PO4, is an ingredient of some preparations used to clean painted walls before they are repainted. What is the pH of 1.0 M Na3PO4? Kb = 2.410-2 PO43- + H2O → OH- + HPO42- Initial concs. 1.0 M 0 M 0 M Changes -x M +x M +x M Equilibrium (1.00 - x) M x M x M Concentration General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 EXAMPLE 17-9 [OH-] [HPO42-] [PO43-] Kb= x · x (1.00 - x) = = 2.410-2 x2 + 0.024x – 0.024 = 0 x = 0.14 M pOH = +0.85 pH = 13.15 It is more difficult to calculate the pH values of NaH2PO4 and Na2HPO4 because two equilibria must be considered simultaneously. General Chemistry: Chapter 17 Prentice-Hall © 2007

Concentrated Solutions of Polyprotic Acids For solutions that are reasonably concentrated (> 0.1 M) the pH values prove to be independent of solution concentrations. for H2PO4- pH = 0.5 (pKa1 + pKa2) = 0.5 (2.15 + 7.20) = 4.68 for HPO42- pH = 0.5 (pKa1 + pKa2) = 0.5 (7.20 + 12.38) = 9.79 General Chemistry: Chapter 17 Prentice-Hall © 2007

17-6 Acid-Base Equilibrium Calculations: A Summary Determine which species are potentially present in solution, and how large their concentrations are likely to be. Identify possible reactions between components and determine their stoichiometry. Identify which equilibrium equations apply to the particular situation and which are most significant. General Chemistry: Chapter 17 Prentice-Hall © 2007

Focus On Buffers in Blood CO2(g) + H2O H2CO3(aq) H2CO3(aq) + H2O(l) HCO3-(aq) Ka1 = 4.410-7 pKa1 = 6.4 pH = 7.4 = 6.4 +1.0 [HCO3-] pH = pKa1 + log [H2CO3] General Chemistry: Chapter 17 Prentice-Hall © 2007

General Chemistry: Chapter 17 Buffers in Blood 10/1 buffer ratio is somewhat outside maximum buffer capacity range but… The need to neutralize excess acid (lactic) is generally greater than the need to neutralize excess base. If additional H2CO3 is needed CO2 from the lungs can be utilized. Other components of the blood (proteins and phosphates) contribute to maintaining blood pH. General Chemistry: Chapter 17 Prentice-Hall © 2007

End of Chapter Questions Don’t waste time making your work pretty. Write what you know to be true down. There are no marks for beauty, just for solutions. Once you have a solution: Consider the final path from start to finish. Review the side paths that terminated. Observe where the critical decision points were. General Chemistry: Chapter 17 Prentice-Hall © 2007